Question

127 127 127 127 127 127​

Answer 1

Ответ:

$4.13.a)\ \ M_1\Big(\dfrac{\pi}{4}\Big):\ \ \dfrac{\pi}{4}+2\pi n\ ,\ n\in Z\\\\b)\ \ 5\ radian=\dfrac{5\cdot 180^\circ }{\pi }\approx 5\cdot 57,3^\circ =286,5^\circ \\\\M_2(5):\ \ 5+2\pi n\ ,\ n\in Z\\\\c)\ \ M_3\Big(\dfrac{3\pi}{4}\Big):\ \ \dfrac{3\pi}{4}+2\pi n\ ,\ n\in Z\\\\d)\ \ -3\ radiana=\dfrac{-3\cdot 180^\circ }{\pi}\approx -3\cdot 57,3^\circ =-171,9^\circ \\\\M_4(-3):\ \ -3+2\pi n\ ,\ n\in Z$

$4.5.-4.11.\\\\\dfrac{\pi}{2}=90^\circ \ \ ,\ \ \ A(0:1)\\\\7\pi =\underbrace{3\cdot 2\pi }_{period}+\pi \ \ ,\ \ \ B(-1;0)\\\\\dfrac{\pi}{3}=60^\circ \ \ ,\ \ \ C(-\frac{1}{2}\ ;\ \frac{\sqrt3}{2}\ )\\\\\dfrac{2\pi }{3}=2\cdot 60^\circ =120^\circ \ ,\ \ \ D(-\frac{1}{2}\ ;\ \frac{\sqrt3}{2}\ )\\\\\\\dfrac{4\pi}{3}=4\cdot 60^\circ =240^\circ \ \ ,\ \ \ E(-\frac{1}{2}\ ;-\frac{\sqrt3}{2}\ )$

$-\dfrac{\pi}{2}=-90^\circ \ \ ,\ \ \ F(0:-1)\\\\\dfrac{25\pi }{4}=25\cdot 45^\circ =1125^\circ =\underbrace{3\cdot 360^\circ }_{period}+45^\circ\ \ ,\ \ \ G(\, \frac{\sqrt2}{2}\ ;\ \frac{\sqrt2}{2}\ )$